UPSC NDA I 2024 Mathematics Paper

To find the value of $\beta$, we need to solve the given integral and compare it with the provided solution form. The integral is:$\int \frac{dx}{\sqrt{x+1}-\sqrt{x-1}}$Let's first simplify the integrand. To do this, multiply the numerator and the denominator by the conjugate of the denominator:$\int \frac{\sqrt{x+1}+\sqrt{x-1}}{(\sqrt{x+1}-\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1})} dx$The denominator simplifies to:$(\sqrt{x+1})^2 - (\sqrt{x-1})^2 = (x+1)-(x-1)=2$So the integral now becomes:$\int \frac{\sqrt{x+1}+\sqrt{x-1}}{2} dx = \frac{1}{2} \int (\sqrt{x+1}+\sqrt{x-1}) dx$This integral can be split into two separate integrals:$\frac{1}{2} \left( \int \sqrt{x+1} dx + \int \sqrt{x-1} dx \right)$For the first integral:$\int \sqrt{x+1} dx$ Let $u = x+1$, hence $du = dx$. Then it becomes:$\int \sqrt{u} du = \frac{2}{3} u^{\frac{3}{2}} = \frac{2}{3} (x+1)^{\frac{3}{2}}$For the second integral:$\int \sqrt{x-1} dx$Let $v = x-1$, hence $dv = dx$. Then it becomes:$\int \sqrt{v} dv = \frac{2}{3} v^{\frac{3}{2}} = \frac{2}{3} (x-1)^{\frac{3}{2}}$Combining these results, we get: $\frac{1}{2} \left( \frac{2}{3} (x+1)^{\frac{3}{2}} + \frac{2}{3} (x-1)^{\frac{3}{2}} \right) = \frac{1}{3} (x+1)^{\frac{3}{2}} + \frac{1}{3} (x-1)^{\frac{3}{2}} + c$Comparing with the given form:$\alpha (x+1)^{\frac{3}{2}} + \beta (x-1)^{\frac{3}{2}} + c$We see that:$\alpha = \frac{1}{3} \quad \text{and} \quad \beta = \frac{1}{3}$